Answered: Z is a standard normal random variable.…

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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Let z be a standard normal random variable. Use the calculator provided, or this table, to determine the value of c. P(-csZsc)=0.9625 Carry your intermediate computations to at least four decimal places. Round your answer to two decimal places. ?
The tread life of tires mounted on light-duty trucks follows the normal probability distribution with a population mean of 60,000 miles and a population standard deviation of 4,000 miles. Suppose we select a sample of 90 tires and use a simulator to determine the tread life. What is the likelihood of finding that the sample mean is between 59,050 and 60,950? 0.4878 0.9756 0.9878 0.5122
Let Z be a standard normal random variable. Use the calculator provided, or this table, to determine the value of c. P(Zsc) =0.1685 Carry your intermediate computations to at least four decimal places. Round your answer to two decimal places.
In a study of life expectancy in a certain region, the mean age at death was 70 years, and the standard deviation was 5 years. The life expectancy (years) is assumed to follow a skewed distribution. Suppose that a random sample of 45 people from this region is selected. The mean life expectancy (years) of the sample is considered a random variable. What is its standard error? 0.5976 O 2.2361 O 5.2341 O 0.1111
You may need to use the appropriate appendix table to answer this question. Given that z is a standard normal random variable, find z for each situation. (Round your answers to two decimal places.) (d) The area to the right of z is 0.1271. (e) The area to the left of z is 0.6700. (f) The area to the right of z is 0.3300.
Let Z be a standard normal random variable. Use the calculator provided, or this table, to determine the value of c. P(1.18
What is the probability that Z is equal to or greater than 0.6? Z represents a standard normal continuous random variable. Answer: The probability that Z is greater than b is 55%. What is b? Z represents a standard normal continuous random variable. Answer:
Find the mean and standard deviation for each binomial random variable: a. n = 60, ππ = .65 (Round your mean value to 2 decimal places and standard deviation to 4 decimal places.) Mean: Standard Deviation: b. n = 98, ππ = .45 (Round your mean value to 2 decimal places and standard deviation to 4 decimal places.) Mean: Standard Deviation: ππ c. n = 50, = .70 (Round your mean value to 2 decimal places and standard deviation to 4 decimal places.) Mean: Standard Deviation:
How were the answers of 0.0344, 0.4656, 0.1470, and 0.1470 attained?
You may need to use the appropriate appendix table to answer this question. Given that z is a standard normal random variable, find z for each situation. (Round your answers to two decimal places.) (a) The area to the left of z is 0.2119. (b) The area between −z and z is 0.9398. (c) The area between −z and z is 0.2052. (d) The area to the left of z is 0.9951. (e) The area to the right of z is 0.5793.
Let Z be a standard normal random variable. Use the calculator provided, or this table, to determine the value of c. P(Zsc)=0.8888 Carry your intermediate computations to at least four decimal places. Round your answer to two decimal places.
You may need to use the appropriate appendix table to answer this question. Given that z is a standard normal random variable, find z for each situation. (Round your answers to two decimal places.) (a) The area to the left of z is 0.2420. (b) The area between −z and z is 0.9232. (c) The area between −z and z is 0.2128. (d) The area to the left of z is 0.9948. (e) The area to the right of z is 0.6554.

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	0.0000
	0.0793
	0.4978
	0.0030

Z is a standard normal random variable. The P (Z=1.41) equals Question 9 options: 0.0000 0.0793 0.4978 0.0030